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Let $a\in\Bbb{R}$. Suppose $f$ is a real valued continuous function on $[a, \infty)$ satisfying that $\lim\limits_{x\to\infty}f(x)=L\in\Bbb{R}$ I need to show:

$f$ is bounded on $[a, \infty)$

Here is what I have so far: if $f$ is bounded on $[a, \infty)$, then there exists a constant $M$ s.t. $|f(x)| \le M$ for all $x \in [a, \infty)$

From the fact that $\lim\limits_{x\to\infty} f(x) = L$, I can say that given $\epsilon > 0$, there is a $\delta$ st $|f(x) - L| < \epsilon$.

If $L=0$, then $|f(x)| < \epsilon$. I can set $M \le \epsilon$, but I don't think this last part makes any sense because there is no way I can claim that $L =0$

Or what if I say that since $|f(x) - L| < \epsilon$, then $|f(x) - L + L| < \epsilon + |L|$. If I let $M = \epsilon + |L|$, then would it work? Please provide me with some hints

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Let be $\varepsilon = 1$. –  Mario De León Urbina Dec 6 '12 at 16:24
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Is $f$ continuous? –  Nameless Dec 6 '12 at 16:24
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The statement is false. If you add the hypothesis that $f$ is continuous, then it is true; you will use continuity in the proof. I cannot understand your argument. You say "there is a $\delta$" but then make no further reference to $\delta$, and then assume without justification that the inequality holds for all $x$. –  Jonas Meyer Dec 6 '12 at 16:24
    
The statement is incorrect. For instance, consider $f(x) = \dfrac1x$. You need to add the assumption that $f(x)$ is continuous on the entire $\mathbb{R}$. –  user17762 Dec 6 '12 at 16:28
    
@Marvis: The domain is $[a,\infty)$, although this was originally different in the title, so that is not quite a counterexample. (Perhaps you meant $a\leq0$, and define $f(0)=0$ or something.) Continuity on $[a,\infty)$ suffices. –  Jonas Meyer Dec 6 '12 at 16:30
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1 Answer 1

up vote 4 down vote accepted

The continuity condition is quite essential, here. Consider the function $$f(x)=\begin{cases}\frac1x & x\neq 0\\0 & x=0.\end{cases}$$

Then $\lim\limits_{x\to\infty}f(x)=0$, but $f$ is certainly not bounded on $[a,\infty)$ for $a=0$.


Set $\epsilon>0$, so that you know there is some $N>a$ such that $|f(x)-L|<\epsilon$ for $x>N$. (Note the difference between what I've put here and your statement with the $\delta$.) As you noted, we can say that $|f(x)|\le|L|+\epsilon$ for certain $x$--in particular, for $x\in(N,\infty)$.

Now, $f$ is continuous on the compact interval $[a,N]$, so is bounded--meaning that there is some $M_0$ such that $|f(x)|\le M_0$ for all $x\in[a,N]$. Thus, putting $M=\max\{|L|+\epsilon,M_0\}$, it follows that $|f(x)|\le M$ for all $x\in[a,\infty)=[a,N]\cup(N,\infty)$.

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I'm sorry, I didn't see your answer from earlier, thanks! I have a question though, why do you change the compact interval $[a,n]$ to $for all $x\in[a, N]$ up there? –  Akaichan Dec 6 '12 at 17:15
    
That was a typo. I'll fix it. –  Cameron Buie Dec 6 '12 at 18:36
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